Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 - 14r + 48}{r^2 + 5r + 6} \times \dfrac{r + 2}{-3r + 18} $
First factor out any common factors. $q = \dfrac{r^2 - 14r + 48}{r^2 + 5r + 6} \times \dfrac{r + 2}{-3(r - 6)} $ Then factor the quadratic expressions. $q = \dfrac {(r - 6)(r - 8)} {(r + 2)(r + 3)} \times \dfrac {r + 2} {-3(r - 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (r - 6)(r - 8) \times (r + 2)} { (r + 2)(r + 3) \times -3(r - 6)} $ $q = \dfrac {(r - 6)(r - 8)(r + 2)} {-3(r + 2)(r + 3)(r - 6)} $ Notice that $(r + 2)$ and $(r - 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {(r - 6)(r - 8)\cancel{(r + 2)}} {-3\cancel{(r + 2)}(r + 3)(r - 6)} $ We are dividing by $r + 2$ , so $r + 2 \neq 0$ Therefore, $r \neq -2$ $q = \dfrac {\cancel{(r - 6)}(r - 8)\cancel{(r + 2)}} {-3\cancel{(r + 2)}(r + 3)\cancel{(r - 6)}} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $q = \dfrac {r - 8} {-3(r + 3)} $ $ q = \dfrac{-(r - 8)}{3(r + 3)}; r \neq -2; r \neq 6 $